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Liquid oscillating in a U-tube

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A U-tube has vertical arms of radii 'r' and '2r', connected by a horizontal tube of length 'l' whose radius increases linearly from r to 2r. The U-tube contains liquid up to a height 'h' in each arm. The liquid is set oscillating, and at a given instant the liquid in the narrower arm is a distance 'y' above the equilibrium level.

    Show that the potential energy of the liquid is given by...

    U = (5/8)gρπ(r2)(y2)

    'y' is the change in height.

    2. Relevant equations
    So i have the potential energy of system like his is given by
    U = gρπ(r2)(y2)

    3. The attempt at a solution

    Im not sure where this 5/8's comes from. Im pretty sure its from the fact that the raduis of the tube is changing from r to 2r. Any hints?
     
  2. jcsd
  3. Oct 7, 2012 #2

    Redbelly98

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    For "Relevant equations", you are supposed to list any equation(s) that you'd use as a starting point for solving the problem. What you wrote above would be part of your "attempt at a solution".

    So, what equation (or equations) have you used as a starting point for this problem?

    Yes, it does have to do with the changing radius.

    Another way to think about it: the scenario described is equivalent to taking some water from the "2r" side of the tube and putting it in the "r" side. I think you have figured out the volume and mass of that amount of water (if not -- do it). What is needed is to calculate a change in height, or something along those lines, for moving the water.
     
  4. Oct 7, 2012 #3
    Ok In class we did a tube that was all the radius
    We found that the m was the density x the Area
    Would that be the same here? But isnt the area different b/c of the different r values?
    And if it pluged that value into p=m/v, is that how I'd find the volume?
     
  5. Oct 8, 2012 #4

    Redbelly98

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    Almost, not quite though. Mass is density x Volume.
    It would be, if you knew both p and m. But we don't know m, so it won't help here.

    Instead, you first need to use geometry to figure out the volume.
     
  6. Oct 8, 2012 #5
    Ok so v= πr2 h
    the h would be the length, I think.
    the r2 has to do with the raduis, but thats changing, doesnt chaning things like that have something to do with the ∫?
     
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